Question

Calculate the pH of $0.1$M solution of (i) $NaHC{O}_3$, (ii) $N{a}_{2}HP{O}_4$ and (iii) $Na{H}_{2}P{O}_4$.

Given that:
$C{O}_2+H_{2}O\rightleftharpoons H^{+}+C{O}_3^{2-}$; $K_1=4.2\times {10}^{-7}$M
$HC{O}_3^{-}\rightleftharpoons H^{+}+C{O}_3^{2-}$; $K_2=4.8\times {10}^{-11}$M
$H_{2}P{O}_4\rightleftharpoons H^{+}+H_{2}P{O}_4^{-}$; $K_1=7.5\times {10}^{-3}$M
$H_{2}P{O}_4^{-}\rightleftharpoons H^{+}+HP{O}_4^{2-}$; $K_2=6.2\times {10}^{-8}$M
$HP{O}_4^{2-}\rightleftharpoons H^{+}+P{O}_4^{3-}$; $K_3=1.0\times {10}^{-12}$M.

Answer 1

i) ${CO}_2+H_{2}O\rightleftharpoons H^{+}+{HCO}_3^{-}$ ; $K_1=4.2\times {10}^{-7}M$

${HCO}_3^{-}\rightleftharpoons H^{+}+{CO}_3^{2-}$ ; $K_2=4.8\times {10}^{-11}M$

$\therefore$ $K_1=\frac{\left[H^{+}\right]\left[{HCO}_3^{-}\right]}{\left[H_2{CO}_3\left(aq\right)\right]}$ and $K_2=\frac{\left[H^{+}\right]\left[{CO}_3^{2-}\right]}{\left[{HCO}_3^{-}\right]}$

$\therefore$ $K_1\times K_2=\frac{\left[H^{+}\right]\left[{CO}_3^{2-}\right]}{\left[H_2{CO}_3\right]}={\left[H^{+}\right]}^2\Rightarrow \left[H^{+}\right]=\sqrt{K_1{K}_2}$

$\therefore$ $pH=\frac{{pK}_1+{pK}_2}{2}=\frac{6.37+10.31}{2}=8.34$

ii) ${Na}_2{HPO}_4$

$pH=\frac{{pK}_2+{pK}_3}{2}=\frac{7.2+12}{2}=9.6$

iii) ${NaH}_2{PO}_4$

$pH=\frac{{pK}_1+{pK}_2}{2}=\frac{7.2+2.12}{2}=4.66$